Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$ Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$,where $I$ is the identity linear transformation.
Show that $T$ is  invertible and that $U=T^{-1}$.

From the assumption we have that $TU=I$ and it follows that $U$ is surjection and $T$ is injection,moreover we have the following lemma:
If $V$ and $W$ are finite vectors spaces over the field $\mathbb F$ such that $\text{dim}(V)=\text{dim}(W)$ and $T:V \to W$ is a linear transformation then the following statements are equivalent:

*

*$T$ is invertible

*$T$ is nonsingular,that is $T( \alpha)=0_w \implies  \alpha=0_v$

*$T$ is surjection

So now from the lemma we know that $T$ has an inverse $T^{-1}$ such that $TT^{-1}=I=TU$,the linear transformation $T$ is well-defined and so $T^{-1}TT^{-1}=T^{-1}TU \implies IT^{-1}=IU \implies T^{-1}=U$.
I used the fact that $T$ is well-defied,but is it really well-defined? since the mapping has not been given explicitly then I think it's kinda meaningless to talk about the well-definedness,however without that I'm not able to finish the proof.
 A: 
From the assumption we have that $TU=I$ and it follows that $U$ is surjection and $V$ is injection,...

That's not true, $U$ is an injection since $$(\forall v_1,v_2 \in V) : \ Uv_1=Uv_2 \ \Rightarrow \  v_1 = T(Uv_1) = T(Uv_2) = v_2$$ and $T$ (not $V$) is a surjection since $(\forall v \in V) \ v = T(Uv)$. Anyway, you can also apply the lemma and continue with the proof.

So now from the lemma we now that $T$ has an inverse $T^{−1}$ such that $TT^{−1}=I=TU$, the linear transformation $T$ is well-defined and so...

I don't see the necessity of saying that $T$ is well-defined. What is a well-defined function for you? In this problem we don't need an explicit formula/rule for $T$, it is just an abstract linear map that we are assuming that has a right inverse. So, if, in the future, you are working in a concrete problem and appears a linear operator defined in a finite-dimensional vector space that has a right inverse, this exercise/result shows you that linear operator is necessarily invertible.
A: Let $m(x)=x^k+a_{k-1}x^{k-1}+\cdots+a_1 x + a_0$ be the minimal polynomial for $T$. Then
$$
       (T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I)T=-a_0 I.
$$
So, $V=T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I$ either satisfies $VT=TV=0$, which would imply that $T$ has a non-trivial null space, or else $a_0\ne 0$, and $T$ is invertible with inverse
$$
           T^{-1}=\frac{1}{a_0}(T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I).
$$
Given that $TU=I$, it must be that $T$ is invertible, and $U=T^{-1}$.
